QJEEM19Ma5665: JEE Main 2019 Mathematics - Class 11 JEEM19Ma5665 Solution: Let . If the eccentricity of the hyperbola = 1 is greater than…

Given hyperbola,

${{{x^2}} \over {{{\cos }^2}\theta }} - {{{y^2}} \over {{{\sin }^2}\theta }} = 1$

here a = cos $\theta$

and b = sin $\theta$

We know, eccentricity of the hyperbola is,

$\sqrt {1 + {{{b^2}} \over {{a^2}}}}$

$\therefore$   Here eccentricity

(e) = $\sqrt {1 + {{{{\sin }^2}\theta } \over {{{\cos }^2}\theta }}}$

Given that,

\sqrt {1 + {{{{\sin }^2}\theta } \over {{{\cos }^2}\theta }}} > 2

$\Rightarrow$    \sqrt {1 + {{\tan }^2}\theta } > 2

$\Rightarrow$    1 + tan² $\theta$ > 4

$\Rightarrow$   tan² $\theta$ > 3

$\Rightarrow$   tan $\theta$ > $\pm \sqrt 3$

As given $\theta$ $\in$ $\left( {0,{\pi \over 2}} \right)$

possible value of tan $\theta$ > $\sqrt 3$

So, $\theta$ can be in the range {\pi \over 3} < \theta < {\pi \over 2}



We know latus ractum (LR) = ${{2{b^2}} \over a}$

$\therefore$   LR = ${{2{{\sin }^2}\theta } \over {\cos \theta }}$

= 2 tan $\theta$ sin $\theta$

We know in the range {\pi \over 3} < \theta < {\pi \over 2} tan $\theta$ and sin $\theta$ both are increasing function.

So, at ${\pi \over 3}$ value of LR will be minimum and at ${\pi \over 2}$ value of LR will be maximum.

$\therefore$   Minimum value of LR = 2tan ${\pi \over 3}$ sin ${\pi \over 3}$

= 2 $\times$ $\sqrt 3 \times {{\sqrt 3 } \over 2}$
= 3

Maximum value of LR = 2tan ${\pi \over 2}$ sin ${\pi \over 2}$

= 2 $\left( \infty \right) \times 1$

= $\infty$

$\therefore$   Interval of LR = (3, $\infty$ )